Question

calculate its
A parallel plate capacitor has a area of 4cm2
and plate Separation of 2mn
capacitene
What is new capacitene if a dielectric
of constant 6.7 complety filled between the plates​

Answer 1

Answer:

capacitance= (K×e°× A)/d

therefore= (6.7 × 8.85× 10^-12× 4×10^-4)/ 2×10-3

capacitance= 118.59×10^-13

capacitance= 1.18 ×10^-11

Answer 2

Given: A parallel plate capacitor has a area of 4 cm^2 and a plate separation of 2mm.

To find: Calculate capacitance and the new capacitance if a dielectric of constant 6.7 is completely filled between the plates

Explanation: Area of plate= 4 cm^2

= 4 * 10^-4 m^2

( since 1 cm^2 = 10^-4 m^2)

Distance between plates= 2 mm

= 2* 10^-3 m

(since 1 mm = 10^-3 m)

Let area be denoted by a and distance be denoted by d.

The formula of capacitance is:

$c = \frac{e \times a}{d}$

where e is permittivity in vacuum.

Using values above:

$c = \frac{8.85 \times {10}^{ - 12} \times 4 \times {10}^{ - 4} }{2 \times {10}^{ - 3} }$

$c = 17.7 \times {10}^{ - 13} farad$

When a dielectric is filled completely, the permittivity increases by a factor which is equal to the dielectric constant of the new material.

New permittivity= ke= 6.7 e

New capacitance is given by:

$c = \frac{k \times e \times a}{d}$

$c = \frac{6.7 \times 8.85 \times {10}^{ - 12} \times 4 \times {10}^{ - 4} }{2 \times {10}^{ - 3} }$

$c = 118.59 \times {10}^{ - 13} farad$

The capacitance without dielectric is $17.7 \times {10}^{ - 13}$ farad and capacitance with dielectric constant 6.7 is $118.59 \times {10}^{ - 13}$

farad.